Question

Let $W_1$ and $W_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2+y^2-10x-24y+153=0$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the centre of a circle that is externally tangent to $W_2$ and internally tangent to $W_1.$ If $m^2=\frac{p}{q},$ where $p$ and $q$ are co-prime, then the value of $(p+q)$ is

Answer 1

$W_1:x^2+y^2+10x-24y-87=0$
$C_1=(-5,12)$ and $r_1=16$
Also, $W_2:x^2+y^2-10x-24y+153=0$
$C_2=(5,12)$ and $r_2=4$


Let the circle $W_3$ has centre $C(h,k)=(h,ah)$ and radius $r.$
Then, $C{C}_2=r+4$
$\Rightarrow \sqrt{(h-5{)}^2+(ah-12{)}^2}=r+4\cdots \left(1\right)$
and $C{C}_1=16-r$
$\Rightarrow \sqrt{(h+5{)}^2+(ah-12{)}^2}=16-r\cdots \left(2\right)$
From equation $\left(1\right)$ and $\left(2\right),$ we get
$\sqrt{(h-5{)}^2+(ah-12{)}^2}=20-\sqrt{(h+5{)}^2+(ah-12{)}^2}$
$\Rightarrow 20+h=2\sqrt{(h+5{)}^2+(ah-12{)}^2}$

Again squaring both sides, we get
$(4a^2+3)h^2-96ah+276=0$
For $h$ to be real, $D\ge 0$
$\Rightarrow (-96a{)}^2-4(4a^2+3\left)\right(276)\ge 0$
$\Rightarrow 96\cdot 2a^2-23(4a^2+3)\ge 0$
$\Rightarrow 100a^2\ge 69$
$\Rightarrow a^2\ge \frac{69}{100}$
$\therefore m=\sqrt{\frac{69}{10}}$
$\Rightarrow m^2=\frac{69}{100}=\frac{p}{q}$
$\Rightarrow p=69$ and $q=100$
$\therefore p+q=169$